Counting squares
Let's take a bunch of squares, and put a big "X" in each one, dividing
each square into four triangular wedges. Then let's take three
colors of ink, say red, blue, and black, and ink in the wedges. How
many different ways are there of inking up the squares?
Well, that's easy. There are four wedges, and each one can be one
of three colors, so the answer is 34 = 81. No, wait,
that's not right, because the two squares below are really the
same:

So we have to decide what counts and what doesn't. If one square can be turned
into the other by a one-quarter clockwise or counterclockwise
rotation, or by a half-turn, then we'll say that the two squares are
the same. So all the squares below are "the same":

What about turning the squares over? Are the two squares to the right
"the same"?
Let's say that the squares are inked
on only one side, so that those two would not considered the
same, even if we decided to allow squares with green wedges. Later on
we will make the decision the other way and see how things change.

OK, so let's see. All four wedges might be the same color, and
there are 3 colors, so there are 3 ways to do that, shown at right.

Or there might only be two colors. In
that case, there might be three wedges of one color and one of
another, there are 6 ways to do that, depending on how we pick the
colors; these six are shown at right.
Or it might be two-and-two.
There are three ways to choose the colors (red-blue, red-black, and
blue-black) and two ways to arrange them: same-colored wedges opposite
each other:

or abutting:
so that's another 2·3 = 6 ways.

If we use all three colors, then two wedges are in one of the colors,
and one wedge in each of the other two colors. The two wedges of the
same color might be adjacent to each other or opposite. In either case, we have three choices for the color of the two wedges that are
the same, after which the colors of the other two wedges are forced.
So that's 3 colorings with two adjacent wedges the same color:
And 3 colorings with two opposite wedges the same color:

So that's a total of 21. Unless I left some out. Actually I
did leave some
out, just to see if you were paying attention. There are really 24,
not 21. (You
can see the full set, including the three I left out.) What a
pain in the ass. Now let's do the same count for four colors. Whee,
fun!

But there is a better way. It's called the Pólya-Burnside
counting lemma. (It's named after George Pólya and William
Burnside. The full Pólya counting theorem is more complex and
more powerful. The limited version in this article is more often
known just as the Burnside lemma. But Burnside doesn't deserve the
credit; it was known much earlier to other mathematicians.)

Let's take a slightly simpler example, and count the squares that have
two colors, say blue and black only. We can easily pick them out from the
list above:

So the counting lemma, whatever it is, should get us a count of six.
Here's how it works.

Remember way back at the beginning where we decided that
and
and
were the same because differences
of a simple rotation didn't count? Well, the first thing you do is
you make a list of all the kinds of motions that "don't
count". In this case, there are four motions:

Rotation clockwise by 90°

Rotation by 180°

Rotation counterclockwise by 90°

Rotation by 0°

That last one is a little odd, perhaps, but we have to include it if
we want the right answer. (The mathematics jargon word for these
motions is "symmetries", but I will continue to call them motions.)

Now we temporarily forget about the complication that says that some
squares are essentially the same as other squares. All squares are
now different. and are now different because they are
colored differently. This is a much simpler point of view. There are
clearly 24 such squares, shown below:

For each of the four motions, we count up the number of squares that
would be unchanged by that kind of motion. For example,
every one of the 16 squares is left unchanged by motion #4,
because motion #4 doesn't actually change anything.

Which of these 16 squares is left unchanged by motion #3, a
counterclockwise quarter-turn? All four wedges would have to be the
same color. Of the 16
possible colorings, only the all-black and all-blue ones are left
entirely unchanged by motion #3. Motion #1, the clockwise
quarter-turn, works the same way; only the 2 solid-colored squares are left
unchanged.

4 colorings are left unchanged by
a 180° rotation. The top wedge and the bottom wedges switch
places, so they must be the same color, and the left and right wedges
change places, so they must be the same color. But the top-and-bottom
wedges need not be the same color as the left-and-right wedges. We
have two independent choices of how to color a square so that it will
remain unchanged by a 180° rotation, and there are 22 =
4 colorings that are left unchanged by a 180° rotation. These are
shown at right.

So we have counted the number of squares left unchanged by each
motion:

Motion

# squaresunchanged

typical example

1

Clockwise quarter turn

2

2

Half turn

4

3

Counterclockwise quarter turn

2

4

No motion

16

Next we take the counts for each motion, add them up, and
average them. That's 2 + 4 + 2 + 16 = 24, and divide by 4
motions, the average is 6.

So now what? Oh, now we're done. The average is the answer. 6,
remember? There are 6 distinguishable squares. And our peculiar
calculation gave us 6. Waaa! Surely that is a coincidence? No, it's
not a coincidence; that is why we have the theorem.

Let's try that again with three colors, which gave us so much trouble
before. We hope it will say 24. There are now 34 basic
squares to consider.

For motions #1 and #3, only completely solid colorings are left
unchanged, and there are 3 solid colorings, one in each color. For
motion 2, there are 32 colorings that are left unchanged,
because we can color the top-and-bottom wedges in any color and then
the left-and-right wedges in any color, so that's 3·3 = 9. And
of course all 34 colorings are left unchanged by motion
#4, because it does nothing.

That was so easy, let's skip doing four colors and jump right to the
general case of N colors:

Motion

# squaresunchanged

typical example

1

Clockwise quarter turn

N

2

Half turn

N2

3

Counterclockwise quarter turn

N

4

No motion

N4

Add them up and divide by 4, and you get (N4 +
N2 + 2N)/4. So if we allow four colors, we
should expect to have 70 different squares. I'm glad we didn't try to
count them
by hand!

(Digression: Since the number of different colorings must be an
integer, this furnishes a proof that
N4 +
N2 + 2N is always a multiple of 4. It's a
pretty heavy proof if it were what we were really after, but as a
freebie it's not too bad.)

One important thing to notice is that each motion of the square
divides the wedges into groups called orbits, which are
groups of wedges that change places only with other wedges in the same
orbit. For example, the 180° rotation divided the wedges into two
orbits of two wedges each: the top and bottom wedges changed places
with each other, so they were in one orbit; the left and right wedges
changed places, so they were in another orbit.
The "do nothing" motion induces four orbits; each wedge is in its own
private orbit. Motions 1 and 3 put all the wedges into a single
orbit; there are no smaller private cliques.

For a motion to leave a square unchanged, all the wedges in each orbit
must be the same color. For example, the 180° rotation leaves a
square unchanged only when the two wedges in the top-bottom orbit are
colored the same and the two wedges in the left-right orbit are
colored the same. Wedges in different orbits can be different colors,
but wedges in the same orbit must be the same color.

Suppose a motion divides the wedges into k orbits. Since there
are Nk ways to color the orbits (N
colors for each of the k orbits), there are
Nk colorings that are left unchanged by the
motion.

Let's try a slightly trickier problem. Let's go back to using 3
colors, and see what happens if we are allowed to flip over the
squares, so that
and are now considered the same.

In addition to the four rotary motions we had before, there are now
four new kinds of motions that don't count:

Motion

# squaresunchanged

typical example

5

Northwest-southeast diagonal reflection

9

6

Northeast-southwest diagonal reflection

9

7

Horizontal reflection

27

8

Vertical reflection

27

The diagonal reflections each have two orbits, and so leave 9 of the
81 squares unchanged. The horizontal and vertical reflections each
have three orbits, and so leave 27 of the 81 squares unchanged. So
the eight magic numbers are 3, 3, 9, and 81, from before, and now the
numbers for the reflections, 9, 9, 27, and 27. The average of these
eight numbers is 168/8 = 21. This is correct. It's almost the
same as the 24 we got earlier, but instead of allowing both
representatives of each pair like , we allow only one, since they are
now considered "the same". There are three such pairs, so this
reduces our count by exactly 3.

Okay, enough squares. Lets do, um, cubes! How many different ways
are there to color the faces of a cube with N colors? Well,
this is a pain in the ass even with the Pólya-Burnside lemma, because
there are 24 motions of the cube. (48 if you allow reflections, but
we won't.) But it's less of a pain in the ass than if one tried to do
it by hand.

This is a pain for two reasons. First, you have to figure out what
the 24 motions of the cube are. Once you know that, you then have to
calculate the number of orbits of each one. If you are a
combinatorics expert, you have already solved the first part and
committed the solution to memory. The rest of the world might have to track
down someone who has already done this—but that is not as hard as it
sounds, since here I am, ready to assist.

Fortunately the 24 motions of the cube are not all entirely different
from each other. They are of only four or five types:

Rotations around an axis that goes through one corner to the
opposite corner.

There are 4 such pairs of vertices, and for each pair, you can turn
the cube either 120° clockwise or 120° counterclockwise. That
makes 8 rotations of this type in total. Each of these motions has
2 orbits. For the example axis above, one orbit contains the top,
front, and left faces and the other contains the back, bottom, and
right faces. So each of these 8 rotations leaves N2
colorings of the cube unchanged.

Rotations by 180° around an axis that goes through the middle
of one edge of the cube and out the middle of the opposite edge.

There are 6 such pairs of edges, so 6 such rotations. Each rotation
divides the six faces into three orbits of two faces each. The one
above exchanges the front and bottom, top and back, and left and right
faces; these three pairs are the three orbits. To be left unchanged
by this rotation, the two faces in each orbit must be the same color.
So N3 colorings of the cube are left fixed by each
of these 6 rotations.

Rotations around an axis that goes through the center
of a face and comes out the center of the opposite face.

There are three such axes. The rotation can be 90° clockwise,
90 ° counterclockwise, or 180°.
The 90° rotations have three orbits. The one shown above puts the
top face into an orbit by itself, the bottom face into another orbit,
and the four faces around the middle into a third orbit. So six of
these nine
rotations leave
N3 colorings unchanged.

The 180° rotations have four orbits. A 180° rotation around
the axis shown above puts the top and bottom faces into private
orbits, as the 90° rotation did, but instead of putting the four
middle faces into a single orbit, the front and back faces go in one
orbit and the left and right into another. Since there are three
axes, there are three motions of the cube that each leave
N4 colorings unchanged.

Finally, there's the "motion" that moves nothing. This motion
leaves every face in a separate orbit, and leaves all
N6 colorings unchanged.

Adding up the contributions of the 24 motions, we get:

8N2 from the vertex rotations

6N3 from the edge rotations

6N3 from the 90° face rotations

3N4 from the 180° face rotations

N6 from the "do nothing" motion

The average of these is (N6
+ 3N4
+ 12N3
+ 8N2) / 24, and this is the number of ways
to color the faces of a cube with N colors. We'd better check
it. If we put in N=1, we get out 1 coloring, which is
obviously correct: if you can paint each face of the cube any color
you want so long as it's black, you are guaranteed to get out an
all-black cube. If we put in N=2, we get out
(64
+ 3·16
+ 12·8
+ 8·4) / 24 = 240/24 = 10 colorings.
And this is in fact the correct answer.

Unfortunately, the Pólya-Burnside technique does not tell you what the
ten colorings actually are; for that you have to do some more work.
But at least the P-B lemma tells you when you have finished doing the
work! If you set about to enumerate ways of painting the faces of the
cube, and you end up with 9, you know you must have missed one. And
it tells you how much toil to expect if you do try to work out the
colorings. 10 is not so many, so let's give it a shot:

6 black and 0 white faces: one cube

5 black and 1 white face: one cube

4 black and 2 white faces: the white faces could be on opposite
sides of the cube, or touching at an edge

3 black and 3 white faces: the white faces could include a pair of
opposite faces and one face in between, or they could be the three
faces that surround a single vertex.

2 black and 4 white faces: the black faces could be on opposite
sides of the cube, or touching at an edge

1 black and 5 white faces: one cube

0 black and 6 white faces: one cube

And that's 1+1+2+2+2+1+1 = 10, as we hoped. With N=3 colors,
there are 57 colorings total, so it's hard to imagine counting them
without making a mistake somewhere. Knowing ahead of time that the
answer is 57 is very helpful.

Care to try it out? There are 4 ways to color the sides of a
triangle with two colors, 10 ways if you use three colors, and
N(N+1)(N+2)/6 if you use N colors.

There are 140 different ways to color a the
squares of a 3×3 square array, counting reflections as
different. If reflected colorings are not counted separately, there
are only 102 colorings. (This means that 38 of the colorings have
some reflective symmetry.) If the two colors are considered
interchangeable (so for example and are considered the same) there are
51 colorings.

You might think it is obvious that allowing an exchange of the two
colors cuts the number of colorings in half from 102 to 51, but it is
not so for 2×2 squares. There are 6 ways to color a 2×2
array, whether or not you count reflections as different; if you
consider the two colors interchangeable then there are 4 colorings,
not 3. Why the difference?